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Given two random variables $X, X'$ that are connected via a transformation $T: X \to X'$, I would like to know if the following is true: $$ \text{cdf}_{X'}(x') = \text{cdf}_X\left(T^{-1}(x')\right). $$ I know that the $\text{pdf}$ of $X'$ is given by $$ \text{pdf}_{X'}(x') = \left|\det \left( \frac{\text{d} \, T^{-1}(x')}{\text{d} \, x'} \right) \right| \, \text{pdf}_X(T^{-1}(x')).$$ According to my (limited) understanding, this together with the change-of-variable theorem implies the correctness of the equation I am not sure about. Am I right or wrong?

Joerg
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    What happens when $T(X)=-X$? If you want a concrete example, let $X$ have a uniform distribution on $[-1,1]$. After you figure out the implications of that, then decide what happens when $T(X)=X^2$. – whuber Apr 06 '16 at 18:01
  • Thank you for replying, could you please elaborate more? I am a bit pressured with time at the moment. – Joerg Apr 06 '16 at 18:11
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    Write down the definition: $F(X'\leq x) = F(T(X)\leq x)$. Then the two are equivalent only when $T$ is monotonically increasing – Alex R. Apr 06 '16 at 18:22
  • I see, I guess that is what whuber was trying to tell me. But in general the identity I was asking about holds if $T$ is invertible, i.e. if $T^{-1}$ exists, right? – Joerg Apr 06 '16 at 19:23

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